$f(x)=-4{{x}^{3}}+3{{x}^{2}}-6x+10$ What is the coefficient for the term containing $(x-3)^2$ in the Taylor polynomial, centered at $x=3$, of $f$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-66$ (Choice B) B $66$ (Choice C) C $-33$ (Choice D) D $33$
Solution: Each term of a Taylor polynomial centered at $~x=3~$ is in the form of $\frac{{{f}^{(n)}}(3){{(x-3)}^{n}}}{n!}\,$. We need the second derivative of $~f\left( x \right)\,$, evaluated at $~x=3\,$. $f(x)=-4{{x}^{3}}+3{{x}^{2}}-6x+10$ $f\,^\prime(x)=-12{{x}^{2}}+6x-6$ $f\,^{\prime\prime}(x)=-24x+6$ Then $f\,^{\prime\prime}(3)=-24\cdot 3+6=-66\,$. Therefore, the coefficient of the term containing $~{{\left( x-3 \right)}^{2}}~$ is $-\frac{66}{2!}=-33\,$.